Question

${lim}_{x\to \frac{x}{3}}\frac{sin(\frac{x}{3}-x)}{2cosx-1}$ is equal to

• A. $\sqrt{3}$
• B. $\frac{1}{2}$
• C. $\frac{1}{\sqrt{3}}$
  
• D. $\sqrt{3}$

Answer 1

The correct option is C

$\frac{1}{\sqrt{3}}$

${lim}_{x\to \frac{x}{3}}\frac{sin(\frac{x}{3}-x)}{2cosx-1}$
$={lim}_{h\to 0}\frac{sin\frac{x}{3}-(\frac{x}{3}-h)}{2cos(\frac{x}{3}-h)-1}$
$={lim}_{h\to 0}\frac{sinh}{2[cos\frac{x}{3}cosh+sin\frac{x}{3}sinh]-1}$
$={lim}_{h\to 0}\frac{sinh}{[\frac{1}{2}cosh+\frac{\sqrt{3}}{2}sinh]-1}$
$={lim}_{h\to 0}\frac{sinh}{cosh+\sqrt{3}sinh-1}$
$={lim}_{h\to 0}\frac{sinh}{-2si{n}^2\frac{h}{2}+\sqrt{3}sinh}$
Dividing $N^T$ and $D^T$ by h :
$={lim}_{h\to 0}\frac{\frac{sinh}{h}}{-(2\times \frac{h}{4})\left(\frac{si{n}^2\frac{h}{2}}{h\times \frac{h}{4}}\right)+\frac{\sqrt{3}sinh}{h}}$
$=\frac{1}{\sqrt{3}}$